Problem: Solve for $x$ and $y$ using elimination. ${2x-6y = -58}$ ${3x+3y = 33}$
Answer: We can eliminate $y$ by adding the equations together when the $y$ coefficients have opposite signs. Multiply the bottom equation by $2$ ${2x-6y = -58}$ $6x+6y = 66$ Add the top and bottom equations together. $8x = 8$ $\dfrac{8x}{{8}} = \dfrac{8}{{8}}$ ${x = 1}$ Now that you know ${x = 1}$ , plug it back into $\thinspace {2x-6y = -58}\thinspace$ to find $y$ ${2}{(1)}{ - 6y = -58}$ $2-6y = -58$ $2{-2} - 6y = -58{-2}$ $-6y = -60$ $\dfrac{-6y}{{-6}} = \dfrac{-60}{{-6}}$ ${y = 10}$ You can also plug ${x = 1}$ into $\thinspace {3x+3y = 33}\thinspace$ and get the same answer for $y$ : ${3}{(1)}{ + 3y = 33}$ ${y = 10}$